Closure of boundary a boundary of closure.

Question

Were ment to show that \begin{equation} \overline {\delta A}=\delta \bar{A} \end{equation} $\bar{A}$ being the closure, $A^\circ$ being the interior and $\delta$A being the boundary. I've tried doing it at such: \begin{align} \overline {\delta A}=\delta A \cup\delta A=\delta A\\ \end{align} \begin{align} \delta\bar{A}=\delta(A\cup\delta A)=\overline {A\cup\delta A}\backslash(AU\delta A)^\circ=A\cup\delta A \backslash A^\circ \end{align} Now the last equal sign must be larger than $\delta A$ since $\delta A$ \ $A^\circ$ is an empty set, I am adding something to $\delta A$, hence it must be larger than $\delta A$. Yet it seems wrong for example if I Drive a set in R$^2$ with a missing line, the inclusion doesn't seem to work. Any help appreciated.

Answer 1

This is false. If $A=\mathbb Q$ then $\delta (A)=\mathbb R$ and $\delta \overline {A} =\emptyset$,